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Now, I am going to apply work using Prescotts on an electrical circuit. *************************** dN = nAv*dt (1)/(2) dq/dt = e*dN/(dN/nAv)
V*I ------------------------- P Notice that the pressure supplied by an EMF has nothing to do with the length of the circuit. A battery hooked to a 1 meter circuit of 1cm² wire uses the same force as a similar battery hooked to a 100 meter circuit of similar wire! Yet, it's obvious that more *work* is being done in the 100 meter circuit than in the 1 meter circuit. The reason why the force is the same while the work isn't is not hard at all to understand. An EMF source creates "electromagnetic pressure" on the anode and/or cathode. Once a circuit is started, this electromagnetic pressure is felt throughout the circuit. You can imagine the electrons as being dominoes. Whether you have 1 meter of "dominoes" falling or 100 meters of "dominoes" falling, the initial pressure or force may be the same, and yet, the amount of work done can be very different. (This obviously means that energy *isn't* conserved. That's right.)
*************************** Now, U is a constant for any given circuit. So, given any circuit, it takes a constant amount of work to move a Coulomb along the circuit. Makes sense that it doesn't vary..
Thus, the rate at which work is done per unit distance depends only on the material. Makes sense..
Each electron gains m_e*2v of energy before it makes a collision and losses it's energy. The collision will take place in t_c seconds. U is the amount of work to move a Coulomb L meters. Thus, in L meters, there will be L/(v*t_c) number of collisions. So, L m_e*2v 2m_e which is correct. --------------------------------------------- |
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